During the reaction of 46 g of acetic acid with 46 g of ethanol in the presence of concentrated sulfuric acid, 54 g

| April 9, 2021 | education

During the reaction of 46 g of acetic acid with 46 g of ethanol in the presence of concentrated sulfuric acid, 54 g of ethyl acetate were formed. Determine the yield of the esterification reaction.

1. CH3COOH + C2H5OH → CH3COOC2H5 + H2O;

2. find the chemical amount of acetic acid:

n (CH3COOH) = m (CH3COOH): M (CH3COOH);

M (CH3COOH) = 12 * 2 + 4 * 1 + 16 * 2 = 60 g / mol;

n (CH3COOH) = 46: 60 = 0.7667 mol;

3.Calculate the chemical amount of ethanol:

n (C2H5OH) = m (C2H5OH): M (C2H5OH);

M (C2H5OH) = 2 * 12 + 6 + 16 = 46 g / mol;

n (C2H5OH) = 46: 46 = 1 mol

4. the amounts of interacting acid and alcohol are related as 1: 1, in further calculations we will use the chemical amount of the deficiency – acetic acid. Let’s find the theoretical mass of ethyl acetate:

ntheor (CH3COOC2H5) = n (CH3COOH) = 0.7667 mol;

mtheor (CH3COOC2H5) = ntheor (CH3COOC2H5) * M (CH3COOC2H5);

M (CH3COOC2H5) = 12 * 4 + 8 + 2 * 16 = 88 g / mol;

mtheor (CH3COOC2H5) = 0.7667 * 88 = 67.47 g;

5.determine the yield of the esterification reaction:

ν (CH3COOC2H5) = mtract (CH3COOC2H5): mtheor (CH3COOC2H5);

ν (CH3COOC2H5) = 54: 67.47 = 0.8 or 80%.

Answer: 80%.



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